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circle has one center

Theorem 1.

Lemma 1.

A center of a circle must lie in its interior.

Proof.

Suppose not. Let a circle have a centerOOO not lying in its interior. If OOO lies on the circle, then the circle is degenerate (just a point). Suppose that OOO lies in its exterior. Choose a point AAA on the circle such that a line that containsAAA and OOOpasses through the interior of the circle (i.e., the line is not tangent to the circle). Such a line intersects the circle at another point BBB. It cannot be the case that OOO is in between AAA and BBB, lest OOO be in the interior of the circle. Without loss of generality, let BBB be between AAA and OOO.

..AAABBBOOO

Then B⁢O<A⁢OBOAOBO<AO, a contradiction, since OOO is supposed to be the center of the circle.
∎

Now to prove the theorem.

Proof.

By definition, every circle has at least one center. Suppose that a circle has more than one center. Let OOO and O′superscriptOnormal-′O^{{\prime}} be two distinct centers of this circle. By the previous lemma, OOO and O′superscriptOnormal-′O^{{\prime}} must lie in the interior of the circle. Draw a chord of the circle which contains both OOO and O′superscriptOnormal-′O^{{\prime}}. Let AAA and BBB be the intersections of this chord with the circle such that OOO is betweenAAA and O′superscriptOnormal-′O^{{\prime}}. Since OOO and O′superscriptOnormal-′O^{{\prime}} are in the interior of the circle, it must be the case that O′superscriptOnormal-′O^{{\prime}} is between OOO and BBB.

..AAAOOOO′superscriptOnormal-′O^{{\prime}}BBB

Note that we must have A≠OAOA\neq O, A≠O′AsuperscriptOnormal-′A\neq O^{{\prime}}, B≠OBOB\neq O, and B≠O′BsuperscriptOnormal-′B\neq O^{{\prime}} as pictured. Otherwise, the circle is degenerate, yielding that O=O′OsuperscriptOnormal-′O=O^{{\prime}}. Because of these four inequalities, we also have that A⁢O>0AO0AO>0, A⁢O′>0AsuperscriptOnormal-′0AO^{{\prime}}>0, B⁢O>0BO0BO>0, and B⁢O′>0BsuperscriptOnormal-′0BO^{{\prime}}>0.

Since OOO is a center of the circle, A⁢O=B⁢OAOBOAO=BO. Since O′superscriptOnormal-′O^{{\prime}} is a center of the circle, A⁢O′=B⁢O′AsuperscriptOnormal-′BsuperscriptOnormal-′AO^{{\prime}}=BO^{{\prime}}. Thus, A⁢O<A⁢O′=B⁢O′<B⁢OAOAsuperscriptOnormal-′BsuperscriptOnormal-′BOAO<AO^{{\prime}}=BO^{{\prime}}<BO, a contradiction. It follows that a circle has exactly one center.
∎

More generally, in Riemannian spaces (and generalizations such as Finsler spaces), this theorem will hold provided that geodesics emanating at a point do not focus at some other point, although they can focus at the same point at which they started. This generalizes what we saw in the case of the sphere and the the projective plane. In both those cases, geodesics focussed but, in the former case, they focussed at the antipodal point but, in the latter case, they only focussed back at their starting point. While on spheres, circles have two centers, we can have Riemannian spaces in which geodesics refocus any number of times, eveninfinitely often, in which cases a circle could have any number of centers, even infinitely many of them. Because, by geodesic deviation, focussing requires positivecurvature, we can assert that circles in spaces of non-positive curvature will have unique centers; for instance, this explains why the result holds in hyperbolic geometry (which has constantnegative curvature).

Along the lines of what we discussed last night, I moved the
comments about what spaces this theorem holds in to the end
in a section on generalization. The reason for this is not to
confuse beginners who may not even have heard of non-Euclidean
geometry whilst, at the same time. presenting more complete
information for the expert.